Question Number 109949 by mnjuly1970 last updated on 26/Aug/20
Answered by mathdave last updated on 26/Aug/20
![solution I=∫_0 ^π ln(((sinx)/(cosx))+((cosx)/(sinx)))dx=∫_0 ^π ln(((sin^2 x+cos^2 x)/(cosxsinx)))dx but cos^2 x+sin^2 x=1 and cosxsinx=((sin2x)/2) I=2∫_0 ^(π/2) ln((1/((sin2x)/2)))dx=2∫_0 ^(π/2) ln((2/(sin2x)))dx ( let y=2x) I=(1/2)∫_0 ^π ln((2/(siny)))dy=ln 2∫_0 ^(π/2) dx−∫_0 ^(π/2) ln(siny)dx but ∫_0 ^(π/2) ln(siny)=−(π/2)ln2 I=ln2[x]_0 ^(π/2) −(−(π/2)ln2)=((π/2)ln2+(π/2)ln2)=πln2 ∵∫_0 ^π ln(tanx+cotx)dx=πln2 Q.E.D](https://www.tinkutara.com/question/Q109962.png)
$${solution} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\frac{\mathrm{sin}{x}}{\mathrm{cos}{x}}+\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}\right){dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\frac{\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{cos}{x}\mathrm{sin}{x}}\right){dx} \\ $$$${but}\:\:\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1}\:\:{and}\:\mathrm{cos}{x}\mathrm{sin}{x}=\frac{\mathrm{sin2}{x}}{\mathrm{2}} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{1}}{\frac{\mathrm{sin2}{x}}{\mathrm{2}}}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{sin2}{x}}\right){dx}\:\:\left(\:\:{let}\:{y}=\mathrm{2}{x}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{sin}{y}}\right){dy}=\mathrm{ln}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{y}\right){dx} \\ $$$${but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{y}\right)=−\frac{\pi}{\mathrm{2}}\mathrm{ln2} \\ $$$${I}=\mathrm{ln2}\left[{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left(−\frac{\pi}{\mathrm{2}}\mathrm{ln2}\right)=\left(\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\frac{\pi}{\mathrm{2}}\mathrm{ln2}\right)=\pi\mathrm{ln2} \\ $$$$\because\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{tan}{x}+\mathrm{cot}{x}\right){dx}=\pi\mathrm{ln2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$
Commented by mnjuly1970 last updated on 26/Aug/20
$${very}\:{good}\:{thank}\:{you}\:{so}\:{much} \\ $$$${master}… \\ $$